Student Attendance Record I问题及解法
2017-05-09 16:11
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问题描述:
You are given a string representing an attendance record for a student. The record only contains the following three characters:
'A' : Absent.
'L' : Late.
'P' : Present.
A student could be rewarded if his attendance record doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).
You need to return whether the student could be rewarded according to his attendance record.
示例:
问题分析:
我们只需要遍历字符串,只要字符串中出现的A的次数不超过1,或者L不会连续出现三次及以上,那么,就返回true。
过程详见代码:
class Solution {
public:
bool checkRecord(string s) {
int resA = 0;
for(int i = 0; i < s.length(); i++)
{
if(s[i] == 'A' && ++resA > 1)
{
return false;
}
if(i + 2 < s.length() && s[i] == 'L' && s[i + 1] == 'L' && s[i + 2] == 'L')
{
return false;
}
}
return true;
}
};
You are given a string representing an attendance record for a student. The record only contains the following three characters:
'A' : Absent.
'L' : Late.
'P' : Present.
A student could be rewarded if his attendance record doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).
You need to return whether the student could be rewarded according to his attendance record.
示例:
Input: "PPALLP" Output: True
Input: "PPALLL" Output: False
问题分析:
我们只需要遍历字符串,只要字符串中出现的A的次数不超过1,或者L不会连续出现三次及以上,那么,就返回true。
过程详见代码:
class Solution {
public:
bool checkRecord(string s) {
int resA = 0;
for(int i = 0; i < s.length(); i++)
{
if(s[i] == 'A' && ++resA > 1)
{
return false;
}
if(i + 2 < s.length() && s[i] == 'L' && s[i + 1] == 'L' && s[i + 2] == 'L')
{
return false;
}
}
return true;
}
};
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